Convert 0.000 015 259 to 32 Bit Single Precision IEEE 754 Binary Floating Point Standard, From a Base 10 Decimal Number

0.000 015 259(10) to 32 bit single precision IEEE 754 binary floating point (1 bit for sign, 8 bits for exponent, 23 bits for mantissa) = ?

1. First, convert to the binary (base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.


0(10) =

0(2)


3. Convert to the binary (base 2) the fractional part: 0.000 015 259.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 015 259 × 2 = 0 + 0.000 030 518;
  • 2) 0.000 030 518 × 2 = 0 + 0.000 061 036;
  • 3) 0.000 061 036 × 2 = 0 + 0.000 122 072;
  • 4) 0.000 122 072 × 2 = 0 + 0.000 244 144;
  • 5) 0.000 244 144 × 2 = 0 + 0.000 488 288;
  • 6) 0.000 488 288 × 2 = 0 + 0.000 976 576;
  • 7) 0.000 976 576 × 2 = 0 + 0.001 953 152;
  • 8) 0.001 953 152 × 2 = 0 + 0.003 906 304;
  • 9) 0.003 906 304 × 2 = 0 + 0.007 812 608;
  • 10) 0.007 812 608 × 2 = 0 + 0.015 625 216;
  • 11) 0.015 625 216 × 2 = 0 + 0.031 250 432;
  • 12) 0.031 250 432 × 2 = 0 + 0.062 500 864;
  • 13) 0.062 500 864 × 2 = 0 + 0.125 001 728;
  • 14) 0.125 001 728 × 2 = 0 + 0.250 003 456;
  • 15) 0.250 003 456 × 2 = 0 + 0.500 006 912;
  • 16) 0.500 006 912 × 2 = 1 + 0.000 013 824;
  • 17) 0.000 013 824 × 2 = 0 + 0.000 027 648;
  • 18) 0.000 027 648 × 2 = 0 + 0.000 055 296;
  • 19) 0.000 055 296 × 2 = 0 + 0.000 110 592;
  • 20) 0.000 110 592 × 2 = 0 + 0.000 221 184;
  • 21) 0.000 221 184 × 2 = 0 + 0.000 442 368;
  • 22) 0.000 442 368 × 2 = 0 + 0.000 884 736;
  • 23) 0.000 884 736 × 2 = 0 + 0.001 769 472;
  • 24) 0.001 769 472 × 2 = 0 + 0.003 538 944;
  • 25) 0.003 538 944 × 2 = 0 + 0.007 077 888;
  • 26) 0.007 077 888 × 2 = 0 + 0.014 155 776;
  • 27) 0.014 155 776 × 2 = 0 + 0.028 311 552;
  • 28) 0.028 311 552 × 2 = 0 + 0.056 623 104;
  • 29) 0.056 623 104 × 2 = 0 + 0.113 246 208;
  • 30) 0.113 246 208 × 2 = 0 + 0.226 492 416;
  • 31) 0.226 492 416 × 2 = 0 + 0.452 984 832;
  • 32) 0.452 984 832 × 2 = 0 + 0.905 969 664;
  • 33) 0.905 969 664 × 2 = 1 + 0.811 939 328;
  • 34) 0.811 939 328 × 2 = 1 + 0.623 878 656;
  • 35) 0.623 878 656 × 2 = 1 + 0.247 757 312;
  • 36) 0.247 757 312 × 2 = 0 + 0.495 514 624;
  • 37) 0.495 514 624 × 2 = 0 + 0.991 029 248;
  • 38) 0.991 029 248 × 2 = 1 + 0.982 058 496;
  • 39) 0.982 058 496 × 2 = 1 + 0.964 116 992;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (losing precision...)


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 015 259(10) =

0.0000 0000 0000 0001 0000 0000 0000 0000 1110 011(2)


5. Positive number before normalization:

0.000 015 259(10) =

0.0000 0000 0000 0001 0000 0000 0000 0000 1110 011(2)


6. Normalize the binary representation of the number.

Shift the decimal mark 16 positions to the right so that only one non zero digit remains to the left of it:


0.000 015 259(10) =

0.0000 0000 0000 0001 0000 0000 0000 0000 1110 011(2) =

0.0000 0000 0000 0001 0000 0000 0000 0000 1110 011(2) × 20 =

1.0000 0000 0000 0000 1110 011(2) × 2-16


7. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point representation:

Sign: 0 (a positive number)

Exponent (unadjusted): -16

Mantissa (not normalized):
1.0000 0000 0000 0000 1110 011


8. Adjust the exponent.

Use the 8 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(8-1) - 1 =

-16 + 2(8-1) - 1 =

(-16 + 127)(10) =

111(10)


9. Convert the adjusted exponent from the decimal (base 10) to 8 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 111 ÷ 2 = 55 + 1;
  • 55 ÷ 2 = 27 + 1;
  • 27 ÷ 2 = 13 + 1;
  • 13 ÷ 2 = 6 + 1;
  • 6 ÷ 2 = 3 + 0;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above:


Exponent (adjusted) =

111(10) =

0110 1111(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 23 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 000 0000 0000 0000 0111 0011 =

000 0000 0000 0000 0111 0011


12. The three elements that make up the number's 32 bit single precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (8 bits) =
0110 1111

Mantissa (23 bits) =
000 0000 0000 0000 0111 0011


Number 0.000 015 259 converted from decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point:
0 - 0110 1111 - 000 0000 0000 0000 0111 0011

(32 bits IEEE 754)

More operations of this kind:

0.000 015 258 = ? ... 0.000 015 26 = ?


Convert to 32 bit single precision IEEE 754 binary floating point standard

A number in 32 bit single precision IEEE 754 binary floating point standard representation requires three building elements: sign (it takes 1 bit and it's either 0 for positive or 1 for negative numbers), exponent (8 bits) and mantissa (23 bits)

Latest decimal numbers converted from base ten to 32 bit single precision IEEE 754 floating point binary standard representation

0.000 015 259 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:10 UTC (GMT)
3 000 000 000 683 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
-10.61 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
0.002 329 416 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
15.215 8 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
55.96 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
11 000 101 010 010 001 010 001 000 000 018 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:09 UTC (GMT)
4.343 74 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:08 UTC (GMT)
0.001 012 2 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:08 UTC (GMT)
511.470 100 06 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:08 UTC (GMT)
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21 795 to 32 bit single precision IEEE 754 binary floating point = ? Mar 24 10:08 UTC (GMT)
All base ten decimal numbers converted to 32 bit single precision IEEE 754 binary floating point

How to convert decimal numbers from base ten to 32 bit single precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 32 bit single precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the base ten positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, by shifting the decimal point (or if you prefer, the decimal mark) "n" positions either to the left or to the right, so that only one non zero digit remains to the left of the decimal point.
  • 7. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign if the case) and adjust its length to 23 bits, either by removing the excess bits from the right (losing precision...) or by adding extra '0' bits to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -25.347 from decimal system (base ten) to 32 bit single precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-25.347| = 25.347

  • 2. First convert the integer part, 25. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 25 ÷ 2 = 12 + 1;
    • 12 ÷ 2 = 6 + 0;
    • 6 ÷ 2 = 3 + 0;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    25(10) = 1 1001(2)

  • 4. Then convert the fractional part, 0.347. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.347 × 2 = 0 + 0.694;
    • 2) 0.694 × 2 = 1 + 0.388;
    • 3) 0.388 × 2 = 0 + 0.776;
    • 4) 0.776 × 2 = 1 + 0.552;
    • 5) 0.552 × 2 = 1 + 0.104;
    • 6) 0.104 × 2 = 0 + 0.208;
    • 7) 0.208 × 2 = 0 + 0.416;
    • 8) 0.416 × 2 = 0 + 0.832;
    • 9) 0.832 × 2 = 1 + 0.664;
    • 10) 0.664 × 2 = 1 + 0.328;
    • 11) 0.328 × 2 = 0 + 0.656;
    • 12) 0.656 × 2 = 1 + 0.312;
    • 13) 0.312 × 2 = 0 + 0.624;
    • 14) 0.624 × 2 = 1 + 0.248;
    • 15) 0.248 × 2 = 0 + 0.496;
    • 16) 0.496 × 2 = 0 + 0.992;
    • 17) 0.992 × 2 = 1 + 0.984;
    • 18) 0.984 × 2 = 1 + 0.968;
    • 19) 0.968 × 2 = 1 + 0.936;
    • 20) 0.936 × 2 = 1 + 0.872;
    • 21) 0.872 × 2 = 1 + 0.744;
    • 22) 0.744 × 2 = 1 + 0.488;
    • 23) 0.488 × 2 = 0 + 0.976;
    • 24) 0.976 × 2 = 1 + 0.952;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 23) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.347(10) = 0.0101 1000 1101 0100 1111 1101(2)

  • 6. Summarizing - the positive number before normalization:

    25.347(10) = 1 1001.0101 1000 1101 0100 1111 1101(2)

  • 7. Normalize the binary representation of the number, shifting the decimal point 4 positions to the left so that only one non-zero digit stays to the left of the decimal point:

    25.347(10) =
    1 1001.0101 1000 1101 0100 1111 1101(2) =
    1 1001.0101 1000 1101 0100 1111 1101(2) × 20 =
    1.1001 0101 1000 1101 0100 1111 1101(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 32 bit single precision IEEE 754 binary floating point:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

  • 9. Adjust the exponent in 8 bit excess/bias notation and then convert it from decimal (base 10) to 8 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as already demonstrated above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(8-1) - 1 = (4 + 127)(10) = 131(10) =
    1000 0011(2)

  • 10. Normalize the mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal point) and adjust its length to 23 bits, by removing the excess bits from the right (losing precision...):

    Mantissa (not-normalized): 1.1001 0101 1000 1101 0100 1111 1101

    Mantissa (normalized): 100 1010 1100 0110 1010 0111

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 1000 0011

    Mantissa (23 bits) = 100 1010 1100 0110 1010 0111

  • Number -25.347, converted from the decimal system (base 10) to 32 bit single precision IEEE 754 binary floating point =

    1 - 1000 0011 - 100 1010 1100 0110 1010 0111